Operator theory is the branch of functional analysis that focuses on bounded linear operators, but it includes closed operators and nonlinear operators. Operator theory is also concerned with the study of algebras of operators.

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Reciprocal Non-linear Hammerstein integral equation

I came across a problem that looks like a non-linear Hammerstein equation: $$ \displaystyle y(t)= v(t)+\int_{0}^{\infty} \frac{e^{\iota ts}}{y(s)}\mathrm{d}s $$ I tried solving it by collocation ...
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1answer
163 views

For a Hilbert space $\mathcal{H}$, is every bounded linear operator on $\mathcal{H}$ a linear combination of unitary operators?

Let $(\mathcal{H}, (\cdot, \cdot))$ be a Hilbert space, and let $B \in \mathcal{B}(H)$ be a bounded linear operator on $H$. If $\mathcal{H}$ is a complex Hilbert space, then $B$ can be written as a ...
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2answers
136 views

How to show that $e^{tA}=\frac{1}{2\pi i}\int_{\{Re \ \lambda =a\}}e^{\lambda t}(\lambda I-A)^{-1}d\lambda$?

Let $X$ be a Banach space and $A:X\to X$ be a bounded operator. We can show that if $|\lambda|>|A|$ then $\lambda I-A$ is invertible and $$(\lambda ...
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1answer
31 views

characterization of unital Fourier multipliers on $L^\infty(\mathbb{R})$?

Does there exist a characterization of Fourier multipliers $T \colon L^\infty(\mathbb{R}) \to L^\infty(\mathbb{R})$ which are unital, i.e. $T(1)=1$? In the case of the torus $\mathbb{T}$, it is easy ...
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69 views

Sufficient condition for an operator to be compact in Hilbert space of holomorphic function with respect to Gaussion weight (Fock space).

What I read in a book I could not understand, some one please help. Let $\mathcal{F}=\{f:\mathbb{C^n}\rightarrow\mathbb{C}: \text{$f$ is holomorphic and}\int_{\mathbb{C}^n}\lvert ...
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1answer
79 views

Convergence of square root operators

Let $Q_n$ and $Q$ be compact positive and symmetric operators. Let $A_n = {Q_n}^{\frac12}$ and $A=Q^{\frac12}$. Given $Q_n$ converges to $Q$ w.r.t. operator norm. Does $A_n$ converges to $A$? Thanks. ...
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74 views

A possible Corollary of the Fredholm alternative?

Let $H$ be a Hilbert space, $P : H \rightarrow H$ a positive-definite (bounded) operator and $K : H \rightarrow H$ a compact (not necessarily self-adjoint) operator. Let $T = P + K$. In particular, ...
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1answer
33 views

Definitions of hemicontinuity

can anyone see the equivalence or relation between the following two definitions of hemicontinuity that I encountered: Assume that $K$ is a closed, convex subset of Banach space $X$. Let $X^{*}$ be ...
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1answer
89 views

Skew-adjoint differential operator $B$ with spectrum $\sigma(B)=i(-\infty,-1]$

Consider the Hilbert space $X=L^{2}\left(\mathbb{R}^n\right)$ and the Schrödinger operator $A=i\Delta$ defined on the domain $D(A)=H^2(\mathbb{R}^n)$. It is known that the spectrum of $A$ is ...
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1answer
25 views

Problem involving pseudomonotone mappings on Banach space

I have the following question regarding mappings on a Banach space $X$. If anyone has an idea or hint as to how to resolve this question it would appreciated. Let $X$ be a Banach space, $X^{*}$ its ...
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92 views

How to express $e^{yS^2}(f(x))$ in closed form where $\frac{d}{dx}=S$

$$ f(x)+\frac{y.f'(x)}{1!}+\frac{y^2 f^{''}(x)}{2!}+\cdots=e^{yS}(f(x))=f(x+y) \text{ where }\frac{d}{dx}=S$$ is a operator $$ f(x)+\frac{y.f''(x)}{1!}+\frac{y^2 ...
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1answer
40 views

$R\mbox{ is a right multiplier and }R(a)b=a\overset{?}{\implies} A\mbox{ is unital }$

Let $A$ be a $C^*$-algebra, and $R:A\to A$ its right multilplier. Is it true that $$ \exists b\in A\quad \forall a\in A \quad R(a)b=a\qquad $$ implies $A$ is unital. I know this is true if A is a ...
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0answers
19 views

T=T** for densely defined operator T

Suppose that $H_1$ and $H_2$ are Hilbert spaces and $T: H_1 \to H_2$ is a densely defined linear map with closed graph. Show that $T = T^{**}$. (I have shown that such a $T$ has a densely defined ...
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2answers
64 views

Spectra of operators on different spaces

Can the same operator when defined on two different spaces have different spectra? For example and operator defined on $C_0$ and on $\ell_2$?
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1answer
85 views

Spectrum of symmetric, non-selfadjoint operator on Hilbert space

I heard that any (unbounded) densely defined and symmetric operator $A: \text{dom}(A)\subset \mathcal{H} \to \mathcal{H}$, which is not selfadjoint, has $\text{spec}( A )= \mathbb{C}$. $\mathcal{H}$ ...
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1answer
109 views

Spectrum of a finite rank operator

If $ T\in B(H)$ is a finite rank operator, then there are orthonormal vectors $e_1,...,e_n$ and vectors $g_1,...,g_n$ such that $Th=\sum_{i=1}^n (h,e_i )g_i$, then we can easily see that $T$ is ...
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1answer
57 views

Definition of exponential for operators

if I have a self-adjoint operator $T:D(T) \rightarrow L^2$, then I define its unitary exponential operator by $$e^{iT}(f) := \lim_{k \rightarrow \infty} e^{iT_{k}}(f),$$ where $T_k(f):=\frac{1}{2} ...
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0answers
71 views

When does an integral operator belong to the Schatten - von Neumann class in terms of its kernel?

It is well known that an integral operator $X: L^2(\mu)\to L^2(\nu)$ with kernel $k(x, y)$ belongs to the Schatten-von Neumann class $\mathfrak S_2$ if and only if $\int |k(x, y)|^2\, d\mu(x)\, ...
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1answer
41 views

What is the domain of an operator?

There seems to be a lot confusion on this notion of a domain of an operator $D(A)$ where $A$ is an operator. Can someone use a simple example to illustrate exactly what this is? Say, let $A$ be a ...
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53 views

Prove that operator is completely continuous

Let's consider Banach space $\ell^\infty$ of bounded sequences $x = \{ \xi_n\}_{n=1}^\infty$: $$ ||x|| = \sup_{n \in \mathbb N} |\xi_n|. $$ Suppose matrix $||a_{i j}||_1^\infty$ specifies operator $A$ ...
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1answer
35 views

Let $A: X \to X$ be a Fredholm operator, then $Ax=y$ has a solution iff $Ax=0$ implies $x=0$?

Let $X$ be a Banach space and let $A: X \to X$ be a Fredholm operator, then $Ax=y$ has a solution iff ($Ax=0$ implies $x=0$)? I can't see how this is implied by the common definitions of Fredholm ...
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1answer
53 views

Relation between $\epsilon$-pseudospectrum of operators

If $H$ is a Hilbert space and $\sigma_{\epsilon}(T)$ denotes the space of all $\epsilon$-pseudospectrum of the operator $T$ and $S, T\in B(H)$ be such that $TS=ST=0$, why ...
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2answers
53 views

local inverse of polynomial

Is there a possibility to invert a polynomial locally? I've got the following problem, concerning control theory: Imagine a ideal amplifier with a feedback loop: Let firstly A be not dependent on ...
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1answer
26 views

Solution sets/ existence and uniqueness of solutions to $Ku-\lambda u=\int^1_0 \frac{x^2}{1+y^3}u(y)dy-\lambda u(x)=f(x)$

Given $$ Ku-\lambda u=\int^1_0 \frac{x^2}{1+y^3}u(y)dy-\lambda u(x)=f(x) $$ A) For what values of $\lambda$ does there exist a unique solution for all $f\in L^2(0,1)$? B) Find the solution set ...
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121 views

Generalized Leibniz Rule

Leibniz Rule states that, $$(f\cdot g)^{(m)}(x)=\sum_{k=0}^m \binom{m}{k} f^{(m-k)}(x)g^{(k)}(x).$$ Writing this with differentiation denoted by $D$, we might say $$D^m (fg) = \sum_{k=0}^m ...
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3answers
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Prove that if $T=T^*$ and $\sigma(T)=\{\lambda\}$, then $T=\lambda I$

Show that if $T$ is a self adjoint linear operator on a Hilbert space such that the spectrum contains a single point $\lambda$, then $T=\lambda I$. Then, show this is false if $T$ is not self ...
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Ways to calculate the spectrum of an operator

Friends, I am learning some very basic stuff of spectral theory and kind of lost, in some sense. I am trying to find ways to compute the spectra of different operators, when they work and don't work. ...
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1answer
59 views

Maximal subspace on which an operator is bounded

Consider the Banach space $X=C[0,1]$ of real continuous function on $[0,1]$ equipped with the supremum norm. Consider the operator $A:D(A)\to X$, $Af=f'$ for each $f\in D(A)=C^1[0,1]$. We can see that ...
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0answers
32 views

Are all cyclic representations irreducible?

I know that for a representation $\pi$ of a $^*$-algebra $\mathcal{A}$ on a Hilbert Space $\mathcal{H}$, if $\pi$ is irreducible then it is cyclic. Is the reverse implication also valid - i.e. is ...
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2answers
91 views

Strongly continuous semigroup of operators which cannot be extended to a group

Let $X$ be a Banach space. We call a family of bounded operators $(T(t))_{t\in \mathbb{R}}$ a strongly continuous group if it satisfies the properties of the strongly continuous semigroup but for ...
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1answer
44 views

Adjoints in Banach spaces: when does $(A')' = A$?

Let H be a Hilbert space and consider a linear bounded operator $A:H\to H$. The adjoint $A^*\in L(H) $ of A with $\forall x,y\in H:\; \langle Ax,y\rangle_H=\langle x,A^*y\rangle_H$ has the property: ...
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1answer
57 views

completely bounded maps - convergence

Let $x$ be a completely bounded map between operator spaces $W \subset \mathbf{B}(\mathcal{H})$ and $V \subset \mathbf{B}(\mathcal{K})$, where $\mathcal{H}$ and $\mathcal{K}$ are Hilbert spaces, and ...
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2answers
44 views

Spectrum of $Tu=\int^1_0 (x+y)u(y)dy$

Given the operator $$Tu(x)=\int^1_0 (x+y)u(y)dy$$ on $L^2(0,1)$, find the spectrum of $T$. For all eigenvalues, find their multiplicities and the eigenfunctions. The kernel is Hilbert Schmidt ...
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1answer
76 views

Domain of multiplication operator

Edit: This question arose due to a misunderstanding, which has now been resolved. Let $\psi \in L^{2}(\mathbb{R})$ be a continuous function. Let $M_{\psi}$ be the multiplication operator on ...
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1answer
79 views

Is $AA^*$ and $A^*A$ self-adjoint?

if I have a densely defined closed linear operator $A$ and $A^* = -A$(same domain also closed). Is this sufficient that $AA^*$ and $A^*A$ are proper self-adjoint operators, assuming that we can also ...
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1answer
43 views

Restriction of an operator on a Banach space

Let $X$ be a Banach space and $A:D(A)\to X$ be an unbounded linear operator such that for all $\lambda>c$ ($c$ some constant), $(\lambda I-A)^{-1}$ exists and is a bounded operator which satisfies: ...
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1answer
64 views

Domain of densely-defined second derivative operator, and its factorization

Let $$-d_x^2: \{f \in L^2[0,1];f \in AC^1[0,1] , f(0)=f(1)\} \rightarrow L^2[0,1]$$ be the second derivative operator. Here $AC^1[0,1]$ is the space of functions whose first derivative is absolutely ...
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2answers
77 views

Decomposition of a positive semidefinite self-adjoint operator?

If I have a positive semi-definite self-adjoint operator $H:D(H) \rightarrow L^2$, is it true that there is always a decomposition $H=A^* A$ available? If this is true, what can we say about the ...
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1answer
103 views

Inverse of operator is not continuous in Banach spaces

Let $X$ be a Banach space. If $A:X\to X$ is an invertible bounded operator (injective, surjective and continuous), then $A^{-1}$ is also bounded. Now can I have an example of an unbounded operator ...
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decomposition of spectrum in case of a Banach algebra

Let $A$ a complex Banach algebra containing a unit $e$. For $x \in A$, you can define the spectrum $\sigma(x)$ and the resolvent set of $x$. However, if you consider a complex Banach space $X$ and a ...
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220 views

Spectral theory - continuous spectrum

imagine that I have some differential operator $D$ that is defined on an interval $[a,b]$. Now, assume that we take the boundary conditions in such a way that this operator is self-adjoint. Then, I ...
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Compact operators on $L^2(G)$ as a reduced cross product of $C_0(G)$ and $G$.

If any of the terminology is unclear then please don't hesitate to point it out. My question is: is it true that when $G$ is a locally compact second countable group then: \begin{equation*} C_0(G) ...
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1answer
31 views

Evaluation of Operator-Valued Function

Hello all; above is my question! :) I've gone through all the way up to the final "and hence deduce that". Up to this point, the question has been fairly straightforward, but I have no idea how to ...
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56 views

Why does s = z+1?

What exactly is Laplace transform? motivated me to ask why unit function is $1/s$ by Laplace transform and $1/(1-z)$ by Z-transform? Both seem to be integrals of delta-pulse and secondary integration ...
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1answer
40 views

Inverse of an operator on two functions

I have the following operator, defined for two twice-differentiable functions $f,g$: $X(f,g):=\frac{(g')^3+fg'f''+g'(f')^2-ff'g''}{g'f''-f'g''}$ This operator has the following property: A curve ...
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1answer
78 views

Eigenvectors for normal operators and their adjoints

Can someone tell me if this proof is correct? Claim:V is a vector space over the Complex field. $T:V\rightarrow V$ is a normal operator. Then if $v\in V$ is an eigenvector with the eigenvalue ...
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2answers
41 views

Operator norm with $\inf$

Let $T: V \to W$ be a linear operator. The operator norm is defined as $$ \|T\| = \sup_{v\in V: \|v\|_V = 1} \|Tv\|_W$$ Does $$ \|T\|' = \inf_{v\in V: \|v\|_V = 1} \|Tv\|_W$$ define a norm? I ...
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1answer
71 views

Is a linear operator on $\ell^2$ defined by the inner product necessarily bounded? [duplicate]

If $a=\{a_n\}\in \ell^\infty(\mathbb{R})$ and $\langle a,x \rangle<\infty$ for all $x\in \ell^2(\mathbb{R})$, (where $\langle a, x\rangle=\displaystyle \sum_{k=1}^\infty a_kx_k$), then is $a\in ...
2
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2answers
91 views

If $\|Tv\|=\|T^*v\|$ for all $v\in V$, then $T$ is a normal operator

I have solved a question but I am not sure the last step of the question. If someone can verify it that would be great. Let $V$ be a finite dimensional vector space with complex inner product. Let ...
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1answer
54 views

Fredholm index for 1-d Schroedinger operator

if I look at a Schroedinger-operator $-\frac{d^2}{dx^2} +V$ on a compact intervall $[a,b] \subset \mathbb{R}$ and I take boundary conditions that this operator is self-adjoint (for example periodic ...